Fatigue Index

Metal Fatigue and Endurance

Important note.. The information below is for guidance only . Evaluating the fatigue strength to be used for component design should be carried out using validated material information and with careful consideration of all factors relevant to the stress locations. The links below the table provide more detailed information on fatigue design.

Fatigue Life

Cyclic Loading Life Considerations

A component is stressed to some extent in its operating life.   In all loading scenarios it is desirable to design the component to minimise stress concentrations and maximum the strength of the component material using good design practices.

1) Static Loading..

If the component is stressed to a constant stress level for its operating life then fatigue loading design is not appropriate and for ductile materials the stress concentration factors are not important...If the component is brittle e.g. Cast Iron, then the stress concentration factors need to be considered in the design process..Design using the material yield strength and ultimate strength using the appropriate strength formulae and Factors of Safety can be completed

2) Low life Loading -Stress cycles < 10 3 stress cycles over the design life.

This condition is approached in a similar manner to the static loading scenario.   There is a need to review the loading with respect to the material fatigue properties

Approximate values for low life fatigue strength values for steel are provided below

  • Bending S'l = 0,9 Su
  • Axial Loading S'l = 0,75 Su
  • Axial Loading S'l = 0,72 Su

3) Finite life Loading - Stress cycles 10 3 to 10 6 stress cycles over design lifetime

Use S-N (Wohler) curve for relevant material and determine the relevant fatigue stress level at the relevant design life .  If this information is not available then an estimate of the fatigue strength S'f can be made if the Endurance limit and the Low life strength values are available. ref. High life fatigue strength

The fatigue modifying factors must be considered and the stress concentration factors should also be considered.   If the cyclic stress level at at different values over the operating lifetime then it may be appropriate to use the Palmgren-Miner rule see below..

3) Infinite life Loading - Stress cycles >10 6 stress cycles design lifetime
For ferrous metals and titanium alloys the endurance limit may be used S'n.
For non ferrous the fatigue strength limit S'n may be used,with care , as a design material strength (assuming the n-cycles.
used is similar compared to the projected life).  
The S'n value to be modified by the appropriate fatigue modifying factors and the design should apply appropriate stress concentration values and factors of safety..


In actual service, parts are seldom stressed repeatedly at only one stress level and, hence, the problem arises as to the cumulative damage effect of operations at various levels of stress reversal.    Consequently, the linear cumulative damage rule or the Palmgren-Miner rule has come into common usage.   It assumes that the total life of a part may be estimated by merely adding up the percentage of life consumed by each stress cycle.

Thus, if a specimen, stressed at s1, has a life of N 1 cycles, the damage after n 1 cycles at s1 will be n 1 / N 1 of the total damage, D, at failure.   Similarly, for a two stress level test, where the lives at s1 and s2 are, respectively, N 1 and N 2, the corresponding damages, per cycle, being D/N 1 and D/N 2 the total damage at failure becomes:D = D . n 1 / N 1 + D . n 2 / N 2 or 1 = n 1 /N 1 + n 2 / N 2 where n 1 and n 2 are the total number of cycles at s1 and s1, respectively.

For a multi-level test, Palmgren - Miner rule states

Failure if n 1 / N 1 + n 2 / N 2 + n3 / N3...... > 1

Example :
A component is designed for

  • a stress of 360MPa for 8,000 cycles. Life N 1 from S_N curve = 20,000 cycles
  • a stress of 340MPa for 10,000 cycles. Life N 2 from S_N curve = 40,000 cycles
  • a stress of 280MPa for 40,000 cycles. Life N 3 from S_N curve = 200,000 cycles

8,000 / 20,000 + 10,000/40,000 + 40,000 / 200,000 = 0.8 ( This is less than 1) ..The part will probably not fail in fatigue..

Useful Links..
  1. Efatigue.com... Excellent Source of Fatigue data. (Based on US materials )
  2. Fatigue Basics -epi_eng ...Useful set of notes

Fatigue Index

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